Question

$\frac{1}{81}$ to the base $27$ is $\frac{k}{3}$.
the value of $-2k$ is

Answer 1

${\log }_{27}\frac{1}{81}=\frac{\log \frac{1}{81}}{\log 27}$ ...(${\log }_{b}a=\frac{\log a}{\log b}$)
$\therefore \frac{\log 3^{-4}}{\log 3^3}=\frac{-4}{3}$ ...($\log a^b=b\log a$)

$\frac{k}{3}=\frac{-4}{3}$

k=-4

-2k=8